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### Topic: Simple Stoichiometry Question  (Read 13247 times)

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#### Navi00

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##### Simple Stoichiometry Question
« on: January 30, 2006, 09:02:02 PM »
It's real simple, but you have to start somewhere I guess, I just need someone to take me through this step by step once and I should have it down :

<i>Calculate the amounts requested if 1.34 mol H202 completely react according to the following equation.

2H20 ---> 2H20 + O2

Moles of Oxygen Formed ?

Moles of Water Formed ?</i>

Thanks =P

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #1 on: January 30, 2006, 09:21:02 PM »
First work out the mole ratios:

H2O2 : H2O : O2

2 : 2 : 1

or

1 : 1 : 1/2

so if you have 1.34 moles of H2O2:

1.34 : 1.34 : 0.67
There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #2 on: January 30, 2006, 09:27:03 PM »
Hmmm... How exactly did you get those mole ratio?

I see two H atoms two O atoms in H202 but I only see two H atoms and 1 O atoms in H20 so wouldn't it be 1 1/2?

Thank you =P

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #3 on: January 30, 2006, 09:42:16 PM »
According to your reaction equation, 2 moles of hydrogen peroxide react to form 2 moles of water and 1 mole of oxygen.

Quote
I see two H atoms two O atoms in H202 but I only see two H atoms and 1 O atoms in H20 so wouldn't it be 1 1/2?

I don't understand this bit??
There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #4 on: January 30, 2006, 09:46:22 PM »
Erhmm.. just explain how you impliment the mole ratio into the equation. Sorry, having a headache today ><

Thanks =p

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #5 on: January 30, 2006, 10:00:16 PM »

2H2O2 -----> 2H2O + O2

this is saying that for every 2 moles of peroxide that react you will make 2 moles of water and 1 mole of oxygen.

It could also be written as;

H2O2 --------> H2O + 1/2 O2

which would be saying that for every mole of peroxide that decomposes you create 1 mole of water and half a mole of oxygen.

But lets go back to you original equation:

2H2O2 -----> 2H2O + O2

the ratio of peroxide to water to oxygen is:

H2O2 : H2O : O2

2    :  2    :  1

so if you have 1.34 moles of H2O2:

1.34 : 1.34 : 0.67
There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #6 on: January 30, 2006, 10:12:22 PM »
Ahh okay so I just take 1.34 mol H202 x     1 mol 02
----------     =  0.67 of O2
2 mol H202

and

1.34 mol H202 x  2mol H20
-----------   =   1.34 mol of H20
2mol H202

I appreshate the help mike =p

#### Navi00

• Guest
##### Re:Simple Stoichiometry Question
« Reply #7 on: January 30, 2006, 10:19:41 PM »
Here is my next question if you don't mind helping me visulize it a bit better :

Calculate the amounts requested if 3.30 mol Fe203 completely react according to the following equation.

Fe203 + 2Al ---> 2Fe +Al203

a. Moles of Al needed

b. moles of iron formed

c. moles of aluminum oxide formed

(I'm not trying to get out of homework, just trying to get a better mindset on this type of problem =P) Thanks for the help again

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #8 on: January 30, 2006, 10:23:39 PM »
Ok, so use the info I gave you earlier. See if you can write the mole ratio for this one yourself first
There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #9 on: January 30, 2006, 10:34:36 PM »
Alright... lets see =p

3.30 mol Fe2O3

Fe2O3 +2Al ---> 2Fe + Al2O3
1 : 2  :   2    :    2  :   1 : 2

2Al = 27.0*2 = 54.0

54.0mol Al x 2 mol Al
-----------    =  108 mol of Al needed
1 mol Al203

Hmmm.... is that right?

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #10 on: January 30, 2006, 10:43:58 PM »
No.

You are mixing moles and molecular weight and making the answer unnecessarily complicated.

Fe2O3 +2Al ---> 2Fe + Al2O3

Fe2O3 : Al : Fe : Al2O3

1    : 2  : 2  :   1

so 1 mole of iron oxide reacts with 2 moles of aluminium to produce 2 moles of iron and 1 mole of aluminium oxide.

Therefore:

3.30 : 6.60 : 6.60 : 3.30

so 3.30 moles of iron oxide reacts with 6.60 moles of aluminium to produce 6.60 moles of iron and 3.30 mole of aluminium oxide.

Right?

ps don't try this reaction at home

There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #11 on: January 30, 2006, 10:58:42 PM »
Hmm... it seems to be asking for the answers in grams now, but lets see :

Fe2O3 +2Al ----> 2Fe + Al2O3

1        2           2        1

3.30 mol of Fe2O3 x    2mol Al
----------        = 6.60 mol of Al Required
1 mol Al2O3

3.30 mol of Fe2O3    x   2 mol Fe
------------      = 6.60mol of Fe formed
1 mol Fe2O3

3.30 mol Fe2O3    x   2 mol Al
-------------    = 6.60 mol of Al203 formed
1 mol Fe2O3

Is that right? \ =

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #12 on: January 30, 2006, 11:08:59 PM »
Quote
3.30 mol of Fe2O3 x    2mol Al
----------        = 6.60 mol of Al Required
1 mol Al2O3

3.30 mol of Fe2O3    x  2 mol Fe
------------      = 6.60mol of Fe formed
1 mol Fe2O3

3.30 mol Fe2O3    x  2 mol Al
-------------    = 6.60 mol of Al203 formed
1 mol Fe2O3

Why are you setting it out like this? Is this how the book does it? I haven't seen this before. The ratio of iron oxide to aluminium oxide is 1 : 1, you have discovered this yourself already. So your calculation is wrong because you have calculated that 3.30 moles of iron oxide produces 6.60 moles of aluminium oxide, this is a ratio of 1:2 not 1:1!!

If the ratio of reactants is 1 : 2 (ie iron oxide : aluminium) then every 1 mole of iron oxide reacts with 2 moles of aluminium, that's it no calculation!

Fe2O3 :  Al :  Fe :    Al2O3

1    :    2   :       2   :     1

or

3.30 : 6.60 :  6.60  :  3.30                (moles)

that's it, that is the answer you produce 6.60 moles of Fe and 3.30 moles of aluminium oxide.
There is no science without fancy, and no art without facts.

#### Navi00

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##### Re:Simple Stoichiometry Question
« Reply #13 on: January 30, 2006, 11:23:19 PM »
Hmmm... I guess my book is strange, it has them setup all werid.

Alright, than to convert those to grams you'd just do this? :

27.0g Al / 6.60 mol Al = 4.09g

&

102g Al203 / 3.30 Al203  31.0g

#### mike

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##### Re:Simple Stoichiometry Question
« Reply #14 on: January 30, 2006, 11:36:50 PM »
n = m / M

where:

n = number of moles (mol)
m = mass (g)
M = molecular weight (g.mol-1)

Example:

Al

M = 27 g.mol-1
n = 6.60 mol

n = m/M

therefore:

m = nM = 6.60 x 27 = 178.2 g
There is no science without fancy, and no art without facts.